`N_(2)O_(4)(g)` is dissociated to an extent of `20%` at equilibrium pressure of `1.0` atm and `57^(@)C`. Find the percentage of `N_(2)O_(4)` at `0.2` atm and `57^(@)C`.

Explanation

To solve the problem step by step, we will analyze the dissociation of \( N_2O_4 \) and apply the principles of chemical equilibrium.

### Step 1: Understand the Reaction
The dissociation of \( N_2O_4 \) can be represented as:
\[ N_2O_4(g) \rightleftharpoons 2 NO_2(g) \]

### Step 2: Initial Conditions
At equilibrium pressure of \( 1.0 \, \text{atm} \) and \( 57^\circ C \), \( N_2O_4 \) is dissociated to an extent of \( 20\% \).

Let’s assume we start with 1 mole of \( N_2O_4 \):
- Initial moles of \( N_2O_4 \) = 1 mole
- Initial moles of \( NO_2 \) = 0 moles

### Step 3: Degree of Dissociation
Let \( \alpha \) be the degree of dissociation. Given that \( \alpha = 0.2 \) (20%):
- Moles of \( N_2O_4 \) at equilibrium = \( 1 - \alpha = 1 - 0.2 = 0.8 \) moles
- Moles of \( NO_2 \) at equilibrium = \( 2\alpha = 2 \times 0.2 = 0.4 \) moles

### Step 4: Total Moles at Equilibrium
Total moles at equilibrium:
\[ \text{Total moles} = (1 - \alpha) + 2\alpha = 1 + \alpha = 1 + 0.2 = 1.2 \, \text{moles} \]

### Step 5: Calculate Partial Pressures
Using the total pressure \( P_1 = 1.0 \, \text{atm} \):
- Mole fraction of \( N_2O_4 \):
\[ \text{Mole fraction of } N_2O_4 = \frac{0.8}{1.2} = \frac{2}{3} \]
- Mole fraction of \( NO_2 \):
\[ \text{Mole fraction of } NO_2 = \frac{0.4}{1.2} = \frac{1}{3} \]

Now, calculate the partial pressures:
- Partial pressure of \( N_2O_4 \):
\[ P_{N_2O_4} = \text{Mole fraction of } N_2O_4 \times P_1 = \frac{2}{3} \times 1.0 = \frac{2}{3} \, \text{atm} \]
- Partial pressure of \( NO_2 \):
\[ P_{NO_2} = \text{Mole fraction of } NO_2 \times P_1 = \frac{1}{3} \times 1.0 = \frac{1}{3} \, \text{atm} \]

### Step 6: Calculate \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \]
Substituting the values:
\[ K_p = \frac{\left(\frac{1}{3}\right)^2}{\frac{2}{3}} = \frac{\frac{1}{9}}{\frac{2}{3}} = \frac{1}{9} \times \frac{3}{2} = \frac{1}{6} = 0.1667 \]

### Step 7: New Conditions at \( 0.2 \, \text{atm} \)
Now, we need to find the percentage of \( N_2O_4 \) at a new pressure \( P_2 = 0.2 \, \text{atm} \).

Using the same \( K_p \):
\[ K_p = \frac{4\alpha^2}{1 - \alpha^2} \times P_2 \]
Substituting known values:
\[ 0.1667 = \frac{4\alpha^2}{1 - \alpha^2} \times 0.2 \]

### Step 8: Rearranging the Equation
Rearranging gives:
\[ 0.1667 = \frac{0.8\alpha^2}{1 - \alpha^2} \]
\[ 0.1667(1 - \alpha^2) = 0.8\alpha^2 \]
\[ 0.1667 - 0.1667\alpha^2 = 0.8\alpha^2 \]
\[ 0.1667 = 0.9667\alpha^2 \]
\[ \alpha^2 = \frac{0.1667}{0.9667} \approx 0.172 \]

### Step 9: Solve for \( \alpha \)
Taking the square root:
\[ \alpha \approx 0.414 \]

### Step 10: Calculate Percentage of \( N_2O_4 \)
Finally, the percentage of \( N_2O_4 \) is:
\[ \text{Percentage of } N_2O_4 = (1 - \alpha) \times 100 = (1 - 0.414) \times 100 \approx 58.6\% \]

### Final Answer
The percentage of \( N_2O_4 \) at \( 0.2 \, \text{atm} \) and \( 57^\circ C \) is approximately **58.6%**.